How can one slow down emissions form Flux.interval?

Question

Would this need back pressure or is there a simpler way?

For example in the below code , I want the spin function to be called every 2 seconds. Sometimes 'spin' can take longer time to compute than 2 second interval, in which case I do not want any interval emissions to queue up. But in the below code they do queue up.

In the code below, the first 4 spin function calls take 10 seconds and the rest take 1 second. As a result the Flux.interval emissions 'catch up' once the function gets faster. However, I do not want any 'catch up' to happen

import reactor.core.publisher.Flux;

import java.time.Duration;
import java.util.Date;
import java.util.Iterator;

public class Test {
   public static void main(String[] args) {
        Iterator<Integer> secs = new Iterator<Integer>() {
            private int num = 0;
            @Override
            public boolean hasNext() {
                return true;
            }
            @Override
            public Integer next() {
                return num++ < 4 ? 10 : 1;
            }
        };

        Flux.interval(Duration.ofSeconds(5))
                .map(n -> {spin(secs.next()); return n;})
                .doOnNext(n -> log("Processed " + n))
                .blockLast();

    }

    private static void spin(int secs) {
        log("Current job will take " + secs + " secs");
        long sleepTime = secs*1000000000L; // convert to nanos
        long startTime = System.nanoTime();
        while ((System.nanoTime() - startTime) < sleepTime) {}
    }

    static void log(Object label) {
        System.out.println((new Date()).toString() + "\t| " +Thread.currentThread().getName()   + "\t| " + label);
    }
}

Output: Notice the "Processed" timestamp initially is spaced by 10 seconds, but from job 4 to job 8, there is a 'catch up' that I do not want to take place. I want to spin to executed no earlier than 2 seconds after the previous invocation

Thu Jun 01 17:16:23 EDT 2017    | parallel-1    | Current job will take 10 secs
Thu Jun 01 17:16:33 EDT 2017    | parallel-1    | Processed 0
Thu Jun 01 17:16:33 EDT 2017    | parallel-1    | Current job will take 10 secs
Thu Jun 01 17:16:43 EDT 2017    | parallel-1    | Processed 1
Thu Jun 01 17:16:43 EDT 2017    | parallel-1    | Current job will take 10 secs
Thu Jun 01 17:16:53 EDT 2017    | parallel-1    | Processed 2
Thu Jun 01 17:16:53 EDT 2017    | parallel-1    | Current job will take 10 secs
Thu Jun 01 17:17:03 EDT 2017    | parallel-1    | Processed 3
Thu Jun 01 17:17:03 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:04 EDT 2017    | parallel-1    | Processed 4
Thu Jun 01 17:17:04 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:05 EDT 2017    | parallel-1    | Processed 5
Thu Jun 01 17:17:05 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:06 EDT 2017    | parallel-1    | Processed 6
Thu Jun 01 17:17:06 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:07 EDT 2017    | parallel-1    | Processed 7
Thu Jun 01 17:17:07 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:08 EDT 2017    | parallel-1    | Processed 8
Thu Jun 01 17:17:08 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:09 EDT 2017    | parallel-1    | Processed 9
Thu Jun 01 17:17:13 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:14 EDT 2017    | parallel-1    | Processed 10
Thu Jun 01 17:17:18 EDT 2017    | parallel-1    | Current job will take 1 secs
Thu Jun 01 17:17:19 EDT 2017    | parallel-1    | Processed 11

Answer 1

You can delay events with the delayElements(Duration delay) method. If you want more finegrained control you can use delayUntil(Function<? super T,? extends Publisher<?>> triggerProvider)

Depending on your exact requirements the result might look like this:

 Flux.interval(Duration.ofSeconds(5))
     .map(n -> {spin(secs.next()); return n;})
     .doOnNext(n -> log("Processed " + n))
     .delayUntil(n -> Mono.just(1).delayElements(Duration.Seconds(Math.max(0, 2 - n)))
     .blockLast();